Andres: why not post the answer as an answer: every countable nontrivial partial order is forcing equivalent to the forcing to add a single Cohen real (whose Boolean algebra has size continuum).
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Joel David HamkinsNov 20 '10 at 18:34

I realized my first question had a quick and easy answer shortly after I posted it, so I half expected the same to be true of the second. Thanks!
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Amit Kumar GuptaNov 20 '10 at 18:36

Andres, you could post your answer as an answer if you wanted and I'd click the check mark, but I almost feel like the question was too easy and not worth asking on MO in the first place, so you could delete if you thought that's more appropriate. It was actually just a side question to a bigger question about dominating reals, which I'll post in just a minute.
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Amit Kumar GuptaNov 20 '10 at 18:40

@Amit: Sure. I was about to leave for breakfast, and we just got back. I'll post the answer now.
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Andres CaicedoNov 20 '10 at 19:31

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You don't even need the fact that your forcing has to be Cohen forcing. Being separative, it embeds into its regular open Boolean algebra. In that algebra, each element is the join of some subset of the countable dense set. So the regular open algebra has at most the cardinality of the continuum. More generally, a separative partial order with a dense subset of (infinite) cardinality $\kappa$ must have cardinality at most $2^\kappa$. (Even my use of the regular open algebra seems to be overkill; you could argue directly from separativity.)
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Andreas BlassNov 20 '10 at 19:35

2 Answers
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If ${\mathbb P}$ is a non-trivial separative partial order, and it is countable, an easy argument (back-and-forth) shows that it is forcing isomorphic to Cohen forcing. (This is an exercise in Chapter VII of Kunen's book, I believe, and it can be found in a few other sources, such as the appropriate chapter of the "Handbook of Boolean Algebras".)

It follows that if ${\mathbb P}$ is non-trivial, separative, and admits a countable dense subset, then ${\mathbb P}$ is again Cohen forcing. Now, ${\mathbb P}$ embeds into its Boolean completion which must then coincide with the Boolean completion of Cohen (Borel sets/meager) and therefore has size ${\mathfrak c}=|{\mathbb R}|$. This gives ${\mathfrak c}$ as an upper bound for $|{\mathbb P}|$.

Here is a cute application: Fix $\epsilon>0$, and let ${\mathbb P}^\epsilon$ be the collection of open subsets $p$ of ${\mathbb R}$ that are a finite union of intervals with rational end-points and such that the sum of their lengths (the Lebesgue measure of $p$) is less than $\epsilon$. The generic object gives us an open set that covers the ground model reals, and has measure $\epsilon$. Since ${\mathbb P}^\epsilon$ is countable, it is Cohen forcing. Since $\epsilon$ is arbitrary, it follows that adding a Cohen real makes the set of ground model reals have measure zero.

Edit: As Andreas Blass pointed out, one can actually prove the upper bound $|{\mathbb P}|\le{\mathfrak c}$ directly. This actually gives us a way of proving the characterization of Cohen forcing.

Here's an answer due to Andy Voellmer: If $\mathbb{P}$ has a dense subset $\{ p_i : i < \kappa \}$ of size at most $\kappa$ then by separativity, the map $f : \mathbb{P} \to 2 ^{\kappa}$ defined by $f(p) = \{ i < \kappa : p_i \leq p \}$ is an injection. This is probably what Andreas Blass meant by "arguing directly from separativity."